Problem: This is the easy version of the problem. The difference between the versions is that in the easy version all prices a_i are different. You can make hacks if and only if you solved both versions of the problem.

Today is Sage's birthday, and she will go shopping to buy ice spheres. All n ice spheres are placed in a row and they are numbered from 1 to n from left to right. Each ice sphere has a positive integer price. In this version all prices are different.

An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.

You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.

Input

The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of ice spheres in the shop.

The second line contains n different integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the prices of ice spheres.

Output

In the first line print the maximum number of ice spheres that Sage can buy.

In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.

Example

Input


5
1 2 3 4 5


Output


2
3 1 4 2 5 

Note

In the example it's not possible to place ice spheres in any order so that Sage would buy 3 of them. If the ice spheres are placed like this (3, 1, 4, 2, 5), then Sage will buy two spheres: one for 1 and one for 2, because they are cheap.

Solution (in PYTHON):

Solution:
from __future__ import division, print_function

MOD = 998244353
mod = 10**9 + 7

# import resource
# resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])

def prepare_factorial():
    fact = [1]
    for i in range(1, 100005):
        fact.append((fact[-1] * i) % mod)
    ifact = [0] * 100005
    ifact[100004] = pow(fact[100004], mod - 2, mod)
    for i in range(100004, 0, -1):
        ifact[i - 1] = (i * ifact[i]) % mod

    return fact, ifact

# import threading
# threading.stack_size(1<<27)
import sys
# sys.setrecursionlimit(10000)

# from bisect import bisect_left, bisect_right, insort
from math import floor, ceil, sqrt, degrees, atan, pi, log, sin, radians
# from heapq import heappop, heapify, heappush
from collections import Counter, defaultdict, deque
from itertools import permutations


def modinv(n, p):
    return pow(n, p - 2, p)

def ncr(n, r, fact, ifact):    # for using this uncomment the lines calculating fact and ifact
    t = (fact[n] * (ifact[r] * ifact[n-r]) % mod) % mod
    return t

def get_ints(): return map(int, sys.stdin.readline().strip().split())
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def input(): return sys.stdin.readline().strip()

"""*****************************************************************************************"""

def GCD(x, y):
    while y:
        x, y = y, x % y
    return x

def lcm(x, y):
    return (x * y)//(GCD(x, y))

def get_xor(n):
    return [n, 1, n+1, 0][n % 4]

def get_n(P):    # this function returns the maximum n for which Summation(n) <= Sum
    ans = (-1 + sqrt(1 + 8*P))//2
    return ans

""" ********************************************************************************************* """


def main():

    n = int(input())
    Arr = get_array()
    a = 0
    if n&1:
        a = n//2
    else:
        a = (n//2) - 1

    ans = [0]*n
    Arr.sort()

    curr_index = 0

    for i in range(1, n, 2):
        ans[i] = Arr[curr_index]
        curr_index += 1

    for i in range(0, n, 2):
        ans[i] = Arr[curr_index]
        curr_index += 1

    # print(a)
    # print(*ans)

    a = 0
    for i in range(1, n-1):
        if ans[i-1] > ans[i] < ans[i+1]:
            a += 1

    print(a)
    print(*ans)


""" -------- Python 2 and 3 footer by Pajenegod and c1729 ---------"""

py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192

class FastIO(BytesIO):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0, 2), super(FastIO, self).write(s))[0])
        return s

    def read(self):
        while self._fill(): pass
        return super(FastIO, self).read()

    def readline(self):
        while self.newlines == 0:
            s = self._fill();
            self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)


class IOWrapper(IOBase):

    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s: self.buffer.write(s.encode('ascii'))
            self.read = lambda: self.buffer.read().decode('ascii')
            self.readline = lambda: self.buffer.readline().decode('ascii')


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')

# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')

if __name__ == '__main__':
    main()
    # threading.Thread(target=main).start()